A heavy uniform chain lies on a horizontal tabletop. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is

(a) 20% 

(b) 25% 

(c) 35% 

(d) 15%

Solution:

Let m is the mass of the chain of length L. Let x is the length of the chain that is hanging outside the table without sliding. For equilibrium of the chain, the weight of the hanging part must be balanced by the force of friction on the portion on the table.

Friction force = μMg(L-x)/L

Weight of the hanging chain = Mxg/L

For the system to be stable, μMg(L-x)/L = Mxg/L

μ(L-x) = x

μ = 0.25

=> L/x – 1 = 1/0.25 = 4

L/x = 4+1 = 5

=> x/L = ⅕ = 20%

Hence option a is the answer.

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